Compute the semigroup of endomorphisms of an infinite cyclic semigroup

Compute \mathsf{End}(S) where S is an infinite cyclic semigroup.

We showed in this previous exercise that every infinite cyclic semigroup is isomorphic to (\mathbb{N}^+, +). We now prove a lemma.

Lemma: Let \varphi : S \rightarrow T be a semigroup homomorphism and suppose A generates S. Then \varphi is determined uniquely by its behavior on A in the sense that if \psi : S \rightarrow A is a semigroup homomorphism and \psi(a) = \varphi(a) for all a \in A, then \psi = \varphi. Proof: Since A generates S, every element of S has the form \prod a_i where a : n \rightarrow A for some positive natural number n. Now \varphi(\prod a_i) = \prod \varphi(a_i) = \prod \psi(a_i) = \psi(\prod a_i), so that \psi = \varphi. \square

In particular, \mathbb{N}^+ is generated (additively) by 1. So every endomorphism of \mathbb{N}^+ is determined uniquely by its image at 1.

On the other hand, if we define \varphi_k on \mathbb{N}^+ by \varphi_k(a) = ka, we have \varphi_k(a+b) = k(a+b) = ka + kb = \varphi_k(a) + \varphi_k(b). So \varphi_k is an endomorphism of \mathbb{N}^+. Since \varphi_k(1) = k, by Lemma 1, if \varphi is an endomorphism of \mathbb{N}^+ such that \varphi(1) = k, then \varphi = \varphi_k.

Define \Psi : (\mathbb{N}^+, \cdot) \rightarrow \mathsf{End}(\mathbb{N}^+) by \Psi(k) = \varphi_k. Note that, for all a \in \mathbb{N}^+ (additive), we have \Psi(k\ell)(a) = \varphi_{k\ell}(a) = (k\ell) a = k(\ell a) = \varphi_k(\ell a) = \varphi_k(\varphi_\ell(a)) = (\varphi_k \circ \varphi_\ell)(a) = (\Psi(k) \circ \Psi(\ell))(a). So \Psi(k\ell) = \Psi(k) \circ \Psi(\ell) for all k,\ell \in \mathbb{N}^+, and thus \Psi is a semigroup homomorphism. Certainly \Psi is surjective. Now if \Psi(k) = \Psi(\ell), then in particular, \Psi(k)(1) = \Psi(\ell)(1), so that k = \ell. Hence \Psi is injective, and so an isomorphism.

Thus \mathsf{End}((\mathbb{N}^+,+)) \cong (\mathbb{N}^+, \cdot).

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